What does $sin(arccos(4))-cot(arccos(1))$ equal?

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Answer 1 · 2016-09-27T21:25:09

This expression cannot be calculated. See explanation.

The trigonimetric functions $sin$ and $cos$ have range $<-1;1>$ , so there are NO values of $x$ for which $cosx=4$ , so $arccos(4)$ is NOT defined.

Answer 2 · 2016-09-27T21:44:33

This is undefined on two distinct counts:

(1) If dealing with Real valued functions, then $4$ is not in the range of $cos(x)$ , so $arccos(4)$ is undefined.

(2) $cot(arccos(1))$ is always undefined.

As a Real valued function of Reals, the range of the function $cos(x)$ is $[-1, 1]$ , so $arccos(4)$ is undefined.

But...

Note that $e^(i theta) = cos theta + i sin theta$

Hence:

$cos theta = (e^(i theta) + e^(-i theta))/2$

$sin theta = (e^(i theta) - e^(-i theta))/(2i)$

This yields definitions for $cos z$ and $sin z$ for Complex values of $z$ using the formulae:

$cos z = (e^(iz) + e^(-iz))/2$

$sin z = (e^(iz) - e^(-iz))/(2i)$

With these definitions, we find that the Pythagorean identity still holds:

$cos^2 z + sin^2 z = 1" "$ for all $z in CC$

Hence:

$sin(arccos(4)) = sqrt(1-4^2) = sqrt(-15) = sqrt(15)i$

How about $cot(arccos(1))$ ?

Here we still get an undefined value since:

$arccos(1) = 0$ and $cot(0) = cos(0)/sin(0) = 1/0$ which is undefined.

What does sin(arccos(4))-cot(arccos(1))sin(arccos(4))-cot(arccos(1)) equal?